What is the value of $\dfrac{d}{dx}\left(\dfrac{1}{x^4}\right)$ at $x=2$ ?
Explanation: The strategy We can first rewrite the fraction as a negative power of $x$. Then, the derivative can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have the derivative, we can evaluate it at $x=2$. Rewriting the fraction as a negative power $\dfrac{1}{x^4}=x^{-4}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(x^{-4}\right) \\\\ &=-4x^{-4-1} \gray{\text{The power rule}} \\\\ &=-4x^{-5} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\dfrac{1}{x^4}\right)=-4x^{-5}$, which can also be written as $-\dfrac{4}{x^5}$. Now let's plug ${x=2}$ : $\begin{aligned} -\dfrac{4}{({2})^5}&=-\dfrac{4}{32} \\\\ &=-\dfrac18 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{1}{x^4}\right)$ at $x=2$ is $-\dfrac18$.